Find the limit here please help me?

Question

Find the limit when $\lim\limits_{y\to a}\left[\sin\left(\dfrac{1}{2}(y-a)\right)\cdot\tan\left(\dfrac{\pi y}{2a}\right)\right]$,or at least if you cant solve it show me a hint...

Answer 1

If you plug in your limit right away you get an indeterminate form $0*\infty$. Often such problems can be solved using Lhopitals rule, But to do that you need something resembling a fraction. How can you manipulate the equation into a form of a fraction so that you can apply Lhopitals Rule?

Answer 2

Hint: $\displaystyle\lim_{t\to 0} \frac{\sin t}t =1$, use it for both $t=\displaystyle\frac12(y-a)$, and for $t=\pi/2(1-y/a)$, using also $\tan w = \displaystyle\frac{\cos(\pi/2-w)}{\sin(\pi/2-w)}$.

Answer 3

$\lim\limits_{y\to a}\left[\sin\left(\dfrac{1}{2}(y-a)\right)\cdot\tan\left(\dfrac{\pi y}{2a}\right)\right]$

$=\lim\limits_{x\to 0}\left[\sin\left(\dfrac x2\right)\cdot\tan\left(\dfrac{\pi (x-a)}{2a}\right)\right]$ (Putting $x=y-a$ so $x\to 0$ as $y \to a$)

Now, $\tan\left(\dfrac{\pi (x-a)}{2a}\right)=\tan\left(\dfrac{\pi x}{2a}-\frac{\pi}2\right)=-\tan\left(\frac{\pi}2-\dfrac{\pi x}{2a}\right)$ $=-\cot\left(\dfrac{\pi x}{2a}\right)$

So, $\lim\limits_{x\to 0}\left[\sin\left(\dfrac x2\right)\cdot\tan\left(\dfrac{\pi (x-a)}{2a}\right)\right]$ (Putting $x=y-a$ so $x\to 0$ as $y \to a$)

= $- \lim_{x \to 0}\left( \cos\left(\dfrac{\pi x}{2a}\right) \right)$ $\cdot\lim_{x\to 0}\frac{\sin \frac x 2}{ \frac x 2}$ $\cdot \lim_{x \to 0}\left(\frac{\dfrac{\pi x}{2a}}{\sin\left(\dfrac{\pi x}{2a}\right)} \right) $ $\cdot \lim_{x \to 0}\left(\frac x 2\frac{2a}{\pi x}\right)$

Observe that each limit, except the last is $1$ .